Principle deep dive

Modulus / absolute value behaviour

Splitting |·| at its zero. The principle behind the 2023 modulus spike — the broadest cross-chapter reach in the bank at 11 chapters, from limits and derivatives through definite integrals, areas, equations and relations. Includes the disguised form, where √(f²) is really |f|.

questions in the bank
87
tagged HARD
14%
chapter spread
11
worked examples below
4

When to reach for it

The expression contains |·| — or a square root of something squared, which is |·| in disguise.

Why this principle matters

|x| splits at zero: equals x for x ≥ 0, equals −x for x < 0. That single split drives every modulus question in NDA. Left and right limits diverge at the split point; the function is continuous there but not differentiable.

The principle has been on a tear since 2023 — modulus roughly doubled from about 7 questions per paper-set (2017–22) to about 14 (2023–26), and it has held there for four straight sittings. It now reaches 11 chapters, the widest spread of any principle in the bank: Limits & Continuity, Differentiation, Definite and Indefinite Integration, Apps of Integration, Functions, Quadratic Equations, Sets & Relations, Linear Inequalities, App of Derivatives and Probability. The technique is the same everywhere: split at the zero, handle each piece separately, recombine.

Differentiability is where it is tested hardest. f is differentiable at c only if the left and right derivatives both exist AND agree, and |x| is the canonical counter-example to 'continuous ⇒ differentiable' — at 0 the slopes are −1 and +1. Beware the trap in the other direction: x|x| contains a modulus and IS differentiable at 0, because both one-sided derivatives come out to 0.

Learn to spot the disguise. √(x²) is |x|, not x — so −x/√(x²) is really −x/|x|, a sign function. √(1 − sin 2x) is √((sin x − cos x)²) = |sin x − cos x|, and which branch you take depends entirely on the interval you are given: on (0, π/4) cosine wins, on (π/4, π/2) sine does. The bank sets that same expression twice, once to differentiate and once to integrate, with opposite sign resolutions.

4 worked examples from the bank

Each example demonstrates the principle on a real past-year question. Click to reveal the answer, then the solution.

Example 1Limits & ContinuityEASY
Which one of the following is correct regarding lim⁡x→3∣x−3∣x−3\displaystyle\lim_{x\to3}\dfrac{|x-3|}{x-3}?

[Q74 · Sep · 2024]

Example 2DifferentiationEASY
If f(x)=e∣x∣f(x)=e^{|x|}, then which one of the following is correct?

[Q84 · Apr · 2021]

Example 3Limits & ContinuityMODERATE
If f(x)=x2+x+∣x∣xf(x) = \frac{x^2+x+|x|}{x}, then what is lim⁡x→0f(x)\lim_{x\to 0} f(x) equal to?

[Q86 · Sep · 2022]

Example 4Limits & ContinuityHARD
Let f(x)=∣x∣+1f(x)=|x|+1, g(x)=[x]−1g(x)=[x]-1, h(x)=f(x)g(x)h(x)=\dfrac{f(x)}{g(x)}, where [⋅][\cdot] is the greatest integer function. What is lim⁡x→0−h(x)+lim⁡x→0+h(x)\displaystyle\lim_{x\to0^-}h(x)+\lim_{x\to0^+}h(x) equal to?

[Q88 · Apr · 2024]

Variants to recognise

Same principle, different surfaces. Pattern-match these on test day.

  • Piecewise definition of |x|

    |x| = x for x ≥ 0, −x for x < 0. The split point matters; everything else is algebra.

  • Left vs right limit at the split

    lim x→0⁻ |x|/x = −1; lim x→0⁺ |x|/x = +1. Two-sided limit doesn't exist. NDA exploits this.

  • |x| is continuous, not differentiable at 0

    The graph has a corner. Left derivative = −1, right derivative = +1, so f' undefined at x = 0.

  • Hidden modulus: √(f²) = |f|

    A square root of a perfect square is a modulus, never the bare expression. √(x²) = |x|; √(1 − sin 2x) = |sin x − cos x|. The given interval decides the sign.

  • The x|x| trap

    Contains a modulus yet IS differentiable at 0 — both one-sided derivatives are 0. Presence of |·| is not proof of a corner; always test both sides.

Drill every modulus / absolute value behaviour question

87 questions from the bank — paginated, with cart and Word-export support.

Drill the 87 questions

Related principles

Often combined with this one — drill these next if you found the examples above tractable.